Question

Let $\vec a = 2\hat i + \hat j - 2\hat k$, $\vec b = \hat i + \hat j$ and $\vec c = \vec a \times \vec b$. Let $\vec d$ be a vector such that $|\vec d - \vec a| = \sqrt{11}$, $|\vec c \times \vec d| = 3$ and the angle between $\vec c$ and $\vec d$ is $\frac{\pi}{4}$. Then $\vec a \cdot \vec d$ is equal to
 

• A. 1
• B. 3
• C. 11
• D. 0

Hint:

Use vector identities to reduce problems involving magnitudes and angles to simple dot products.

Answer 1

The Correct Option is D

To solve the given problem, we need to find the dot product \(\vec{a} \cdot \vec{d}\). Let's proceed step-by-step:

1. First, calculate \(\vec{c} = \vec{a} \times \vec{b}\).
   • Given \(\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}\) and \(\vec{b} = \hat{i} + \hat{j}\).
   • Calculate the cross product using the determinant formula: \(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix}\).
   • Expanding the determinant, we get:
     • \(\hat{i}(1 \times 0 - (-2) \times 1) = 2\hat{i}\)
     • \(-\hat{j}(2 \times 0 - (-2) \times 1) = 2\hat{j}\)
     • \(+\hat{k}(2 \times 1 - 1 \times 1) = \hat{k}\)
2. Next, we know that \(|\vec{d} - \vec{a}| = \sqrt{11}\).
   • This implies that \((\vec{d} - \vec{a}) \cdot (\vec{d} - \vec{a}) = 11\).
3. The equation \(|\vec{c} \times \vec{d}| = 3\) gives:
   • Since \(|\vec{c} \times \vec{d}| = |\vec{c}||\vec{d}|\sin\left(\frac{\pi}{4}\right)\), and \(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\),
   • The magnitude of \(\vec{c}\) is \(\sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3\).
   • Thus, \(3|\vec{d}|\frac{\sqrt{2}}{2} = 3\) implies that \(|\vec{d}| = \sqrt{2}\).
4. Given \(|\vec{d} - \vec{a}| = \sqrt{11}\) and \(|\vec{d}| = \sqrt{2}\), the vector \(\vec{d}\) also needs to satisfy that \((\vec{d} - \vec{a}) \cdot (\vec{d} - \vec{a}) = 11\).
5. Substitute and solve to find \(\vec{a} \cdot \vec{d}\).
6. We have proven that the correct answer for \(\vec{a} \cdot \vec{d}\), using all these details is \(0\).

Thus, the correct answer is \(0\).